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svlad2 [7]
3 years ago
14

A sample of carbon dioxide has a pressure of 1.2 atm, a volume of

Chemistry
1 answer:
GalinKa [24]3 years ago
3 0

The answer for the following question is mentioned below.

<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>

Explanation:

Given:

Pressure of gas (P) = 1.2 atm

Volume of a gas (V) = 50.0 liters

Temperature (T) =650 K

To calculate:

no of moles present in the gas (n)

We know;

According to the ideal gas equation;

We know;

<u>P × V = n × R × T </u>

where,

P represents pressure of the gas

V represents volume of the gas

n represents no of the moles of a gas

R represents the universal gas constant  

where the value of R is 0.0821 L atm  mole^{-1}  K^-1

T represents the temperature of the gas

As we have to calculate the no of moles of the gas;

n = \frac{P*V}{R*T}

n = \frac{1.2*50.0}{0.0821*650}

n = \frac{60}{53.365}

n = 1.12 moles

<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>

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kkurt [141]

Moles of H₂ are needed to produce 9.33 moles of NH₃ : 13.995

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

The reaction coefficient in a chemical equation shows the mole ratio of the reactants and products

Reaction for the synthesis of ammonia :

N₂+3H₂⇒2NH₃

moles of NH₃ = 9.33

From equation, mol ratio of H₂ : NH₃ = 3 : 2, so mol H₂ :

\tt =\dfrac{3}{2}\times 9.33\\\\=13.995

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10. A 2.36-gram sample of NaHCO3 was completely decomposed in an
Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

  • Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

m_r=m_{NaHCO_3}=2.36 g

We have two products formed, sodium carbonate and carbonic acid. This implies that the total mass of the products is:

m_p=m_{Na_2CO_3}+m_{H_2CO_3}

Apply the law of mass conservation:

m_r=m_p

Substitute the given variables:

m_{NaHCO_3}=m_{Na_2CO_3}+m_{H_2CO_3}

Rearrange for the mass of carbonic acid:

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Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

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y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

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